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Geek Culture / Is there a pythagorean triple in which every integer is odd?

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Posted: 10th Feb 2007 03:33
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My math teacher has a little bonus calender type thing...and well one of the extra credit problems on it was "what is a pythagorean triple in which each integer is odd?"

now, I have made a program to calculate and test for a pythagorean triple in each of those odd numbers...I currently have got it to 20,001, and then I stopped it because I had to shut down my computer(btw I made this in C)...so now giving up on that, is their a list of pythagorean triples(probably up to 100,000)?

also if anyone cares to make a program to try to find one, starting with C being 20,001 and a and b being from 1 to C

I forgot to upload the source when I came to my moms so I don't have the program to calculate it...

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hyrichter
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Posted: 10th Feb 2007 05:57
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There isn't one.

If you think about it for a minute, you'll realize that at least one side will be an even number because of the way the formula works. The sides must be divisible by 3, 4, and 5 to be a pythagorean triple, and only even numbers are divisible by 4.

http://mathworld.wolfram.com/PythagoreanTriple.html

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Posted: 10th Feb 2007 06:17
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hmmm...maybe I should tell my math teacher with this new found knowledge..(I am right, it is impossible..)

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Matt Rock
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Posted: 10th Feb 2007 19:03
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Just say the answer is "i." Like when trying to find the square root of zero, the answer ends up being i, if I remember correctly that is.


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Venge
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Posted: 10th Feb 2007 19:06
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"i" is the square root of -1

we are learning this (again...) in algebra 2

the square root of 0 would be 0

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Matt Rock
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Posted: 10th Feb 2007 19:22
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oh right, -1... ack, sorry hehe.


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Peter H
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Posted: 10th Feb 2007 19:25 Edited at: 10th Feb 2007 19:27
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sqrt(e^(Pi*i))



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MiR
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Posted: 10th Feb 2007 20:06
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Well if i is used up just use j. j=pythagorean triple in which every integer is odd and is the basis of really complex numbers.

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Antidote
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Posted: 10th Feb 2007 20:12
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In my c++ class last year our midterm exam was to find every Isosceles Pythagorean triple where the measure of the hypotenuse was between 1 and 10000. There were several people who actually coded an algorithm that yielded nothing while those of us who really read the program guidelines simply wrote




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Peter H
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Posted: 10th Feb 2007 20:23 Edited at: 10th Feb 2007 20:25
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@Antidote- took me a moment to figure that out roflCOPTERS!

for anyone who is still wondering...


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Matt Rock
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Posted: 10th Feb 2007 23:38
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Mir, "i" stands for "impossible number" if I'm not mistaken. So what's J? "Jam-Master-Flash of all impossible numbers?"


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Peter H
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Posted: 11th Feb 2007 00:26
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i think "i" stands for "imaginary number" (representing sqrt(-1))

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Matt Rock
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Posted: 11th Feb 2007 04:07 Edited at: 11th Feb 2007 04:07
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Yeesh, I'm batting a million here lol.


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mm0zct
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Posted: 11th Feb 2007 04:16
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Quote: "Pythagoreans theorem doesn't apply to Isosceles triangles because you need a right angle in the triangle.
And Isosceles triangles never have one."


could someone please explain why an Isoscelese triangle can't have a right angle

i think you may be thinking of equlateral. Isoscelese just means 2 sides are the same length, nothing to stop there being an angle of 90° between them, just means the other 2 angles are 45°

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hyrichter
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Posted: 11th Feb 2007 04:25
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An isosceles triangle certainly can have a right angle. However, there can never be an isosceles Pythagorean triple triangle, because the ratios of the sides must be a multiple of 3, 4, and 5.

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Peter H
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Posted: 11th Feb 2007 15:11 Edited at: 11th Feb 2007 15:12
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bleh, you're right...

that's what "higher" () level math does to you, makes you forget all the other math you've learned

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Chris K
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Posted: 11th Feb 2007 18:27
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The sides don't have to be a multiple of 3, 4 and 5. What about 5, 12 and 13?

There are an infinite number of pyth. triples.

However, it is true that there are none where all the numbers are odd. This is because a pyth. triple is a^2 + b^2 = c^2.

If all the numbers are odd this is:

odd^2 + odd^2 = odd^2
or
(odd * odd) + (odd * odd) = (odd * odd)

We know that an odd number times by an odd number is an odd number-

odd + odd = odd

But this is wrong because an odd number added to an odd number is always an even number.

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Hobgoblin Lord
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Posted: 11th Feb 2007 18:57 Edited at: 11th Feb 2007 18:57
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Quote: "Yeesh, I'm batting a million here lol"

More like your batting i.



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Matt Rock
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Posted: 11th Feb 2007 19:30
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lol I should have thought of that


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hyrichter
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Posted: 11th Feb 2007 20:31
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Quote: "The sides don't have to be a multiple of 3, 4 and 5. What about 5, 12 and 13?"

I should've explained it better. One side (a or b) will be divisible by 3, one side (a or b) will be divisible by 5, and one side (a, b, or c) will be divisible by 5. So your example of 5, 12, and 13 works because:
5 is divisible by 3 and 5.
12 is divisible by 3 and 4.
So you still get the 3, 4, and 5, even though the hypotenuse can be a prime number.

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Chris K
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Posted: 11th Feb 2007 22:13
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Where's the proof that one side is always divisible by 4? (This is the basis of your argument, but you haven't proven it, you merely stated it - I am pretty sure it is not true).

It is, I believe, possible to have odd a, b and even c. This would break your current logic but I can't think of an example off the top of my head.

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Posted: 11th Feb 2007 22:18
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theirs not one that has even both odd(or at least to 20, 000)

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Chris K
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Posted: 11th Feb 2007 22:25
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Just checked and one side is div. by 3, one by 4, one by 5. Never knew that!

Can't find a proof I would imagine it is not trivial.

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Osiris
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Posted: 12th Feb 2007 03:31 Edited at: 12th Feb 2007 03:34
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Arguing about triangles eh?
<edit>
Whats the source to that program?
</edit>

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Posted: 24th Feb 2007 03:41
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it turns out it didn't exist! I got the bonus points for it! thanks everyone...2 points in exchange for about 2 hours of scratching my head

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hyrichter
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Posted: 24th Feb 2007 04:59 Edited at: 24th Feb 2007 05:02
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Ermm, yeah, didn't we already determine that long ago?

Edit: @ChrisK. I don't have any proof of the 3, 4, 5 divisibility property, other than just what I read on several math websites and wikipedia. I'm sure if you searched google, you could find a proof, though.

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kevil
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Posted: 24th Feb 2007 15:19
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Here's a nice problem with pythagorean triples.
Consider a box of size lxwxh. Is there a box with

l^2 + w^2 = a^2
l^2 + h^2 = b^2
w^2 + h^2 = c^2

in other words with (l,w,a), (l,h,b) and (w,h,c) pythagorean triples.
Even more interesting: is there such a box where also

l^2 + w^2 + h^2 = d^2

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Posted: 24th Feb 2007 15:25
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oooh! ooooh!
Do I get bonus points of some sort!? or perhaps some bisquits!?
lol

don't know much about boxes...yet(were still in 2-D)

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kevil
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Posted: 24th Feb 2007 16:29
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You will definately get bonus points for the second one!
btw, the fact that it forms a box is not important. The only nice thing about seeing it as a box is that it is a special box where the sides have integer lengths, but also the diagonals have integer length.

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Chris K
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Posted: 24th Feb 2007 16:37
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There are two known ones with integer face diagonals I think, but no one's ever found one where the space diagonal is an integer. It's been checked to some obscenely large number...

It's called an Euler Box I think (maybe an Euclid Box). The other one I think is called a Perfect Box.

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kevil
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Posted: 24th Feb 2007 18:53
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yeah, Euler box. I thought the other one is called a perfect cuboid, but I might be wrong.

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